Let \(\mathrm{f}(\mathrm{x})=\sqrt{(\mathrm{x}-1)}+\sqrt{(\mathrm{x}+24-10 \sqrt{(\mathrm{x}-1})) ; 1<\mathrm{x}<26 \text { be real }}\) valued function Then \(f^{\prime}(x)=\) for \(1<\mathrm{x}<26\) is (a) 0 (b) \(\overline{[1 /(\sqrt{1}-1)]}\) (c) \(2 \sqrt{(x-1)-5}\) (d) \(\sqrt{(x-1)+5}\)

The chain rule is a fundamental tool in calculus used to compute the derivative of a composite function. In essence, when you have a function nested within another function, the chain rule allows you to find the rate of change of the entire combination with respect to a variable.

Imagine you have two functions, let's say \( g(x) \) and \( h(x) \) such that our target function \( f(x) = h(g(x)) \). According to the chain rule, the derivative of \( f(x) \) with respect to \( x \) is the product of the derivative of \( h \) with respect to its argument and the derivative of \( g \) with respect to \( x \), which can be written as \( f'(x) = h'(g(x)) \cdot g'(x) \).

When differentiating, it's essential to multiply the outer function's derivative by the inner function's derivative. This process is best understood through practice and applying it to various types of functions to become comfortable with its use.

Differentiating functions involving square roots can initially seem daunting, but it becomes straightforward with practice, especially when utilizing the chain rule. A square root function can be written in the form \( \sqrt{x} \) which is equivalent to \( x^{1/2} \). The derivative of this function, using basic power rules, is \( \frac{1}{2}x^{-1/2} \) or \( \frac{1}{2\sqrt{x}} \).

When the square root contains more than a simple \( x \) term, like \( \sqrt{g(x)} \), you treat \( g(x) \) as the inner function and apply the chain rule as explained previously. For our exercise example, differentiating the term \( \sqrt{x - 1} \) involved applying this rule alongside the derivative of the inner function \( x - 1 \), leading to the derivative of \( \frac{1}{2\sqrt{x-1}} \).

Understanding how to apply derivatives to square root functions is crucial while dealing with more complex equations, such as those involving nested roots or additional operations within the square root.

Mathematical functions are the building blocks of calculus and provide the groundwork upon which concepts like differentiation are built. A function \( f(x) \) is a relation where each input \( x \) is associated with exactly one output, often called \( f(x) \). Functions can take many forms, including linear, quadratic, polynomial, and transcendental expressions.

When working with functions in calculus, it's vital to recognize the type of function you're dealing with—whether it's a simple linear function or a more complex nested function—since this knowledge dictates the differentiation techniques you'll employ. For example, the function presented in our problem involves square roots, which are a type of radical function. It's not just a matter of finding the derivative of individual terms; understanding their interplay is crucial.

Moreover, exercises in differentiation often involve manipulating a given function into a form that simplifies the differentiation process. By deeply understanding how functions work and how they can be transformed, students are better equipped to handle the intricacies of calculus problems.